Question 3.4.2: Additional Answer Options

2021 - May/June - Paper 1 - Question 3

Or

Option 2

Upwards as positive:

\Delta y=v_{i}\Delta t+1/2a\Delta t^2

=(10)(1,02)+1/2(-9,8)(1,02)^2

\Delta y=5,1\mathrm{~m}(5,102)

h=5,1+3

=8,1\mathrm{m}(8,102)

Or

Downwards as positive:

\Delta y=v_{j}\Delta t+1/2a\Delta t^2

=(-10)(1,02)+1/2(9,8)(1,02)^2

\Delta y=-5,1m(-5,102)

h=5,1+3

=8,1\mathrm{m}(8,102)

Or

Option 3

Upwards as positive:

\Delta y=\left(\frac{v_{i}+v_{f}}{2}\right)\Delta t

=\left(\frac{0-10}{2}\right)(1,02)

\Delta y=-5,1\mathrm{~m}

h=5,1+3

=8,1\mathrm{~m}

Or

Downwards as positive:

\Delta y=\left(\frac{v_{i}+v_{f}}{2}\right)\Delta t

=\left(\frac{0+10}{2}\right)(1,02)

\Delta y=5,1\mathrm{~m}

h=5,1+3

=8,1\mathrm{~m}

Or

Option 4

Upwards as positive:

\Delta y=v_{i}\Delta t+1/2a\Delta t^2

=0+1/2(-9,8)(1,02)^2

\Delta y=-5,1m(-5,098)

\mathrm{h}=5,1+3

=8,1\mathrm{m}(8,098)

Or

Downwards as positive:

\Delta y=v_{i}\Delta t+1/2a\Delta t^2

=0+1/2(9,8)(1,02)^2

\Delta y=5,1m(5,098)

h=5,1+3

=8,1\mathrm{m}(8,098)

Or

Option 5

\left.\begin{array}{l}\mathrm{W}_{\text{net }}=\Delta\mathrm{E}_{\mathrm{k}}\\ \mathrm{w}\Delta\mathrm{y}\cos180^{\circ}=1/2\mathrm{m}\left(\mathrm{v}_{\mathrm{f}}^2-\mathrm{v}_{\mathrm{i}}^2\right)\end{array}\right\}\text{ Any one }

(0,06)(9,8)(\Delta y)\cos180^{\circ}=1/2(0,06)\left(0-10^2\right)

\Delta y=5,1\mathrm{~m}

h=5,1+3

=8,1\mathrm{~m}

Or

Option 6

\left.\begin{array}{l}\left(\mathrm{E}_{\text{mech }}\right)_{3m(\text{ ref/verw })}=\left(\mathrm{E}_{\text{mech }}\right)_{\text{Top }/\text{ Bo }}\\ \left(\mathrm{E}_{\mathrm{P}}+\mathrm{E}_{\mathrm{K}}\right)_{3m(\text{ ref/verw })}=\left(\mathrm{E}_{\mathrm{P}}+\mathrm{E}_{\mathrm{K}}\right)_{Top/Bo}\\ \left(\mathrm{mgh}+1/2\mathrm{mv}^2\right)_{3m(\text{ ref/verw })}=\left(\mathrm{mgh}+1/2\mathrm{mv}^2\right)_{Top/Bo}\end{array}\right\}\text{ Any one }

(0)+1/2(0,06)(10)^2=(0,06)(9,8)(h)+0

\mathrm{h}=5,1\mathrm{~m}

h=5,1+3

=8,1\mathrm{~m}

Or

Option 7

Upwards as positive:

\Delta y=v_{i}\Delta t+1/2a\Delta t^2

-3=10\Delta t+1/2(-9,8)\Delta t^2

\Delta\mathrm{t}=2,31\mathrm{~s}(2,306)

\Delta y=v_{i}\Delta t+1/2a\Delta t^2

=0+1/2(-9,8)(2,31-1,02)^2

\Delta y=-8,15m(8,1)

\mathrm{h}=8,15\mathrm{~m}(8,1)

Or

Downwards as positive:

\Delta y=v_{i}\Delta t+1/2a\Delta t^2

+3=-10\Delta t+1/2(9,8)\Delta t^2

\Delta\mathrm{t}=2,31\mathrm{~s}(2,306)

\Delta y=v_{i}\Delta t+1/2a\Delta t^2

=0+1/2(9,8)(2,31-1,02)^2

\Delta y=8,15\mathrm{~m}(8,1)

\mathrm{h}=8,15\mathrm{~m}(8,1)

Or

Option 8

\left.\begin{array}{l}\left(E_{\text{mech }}\right)_{3m}m(\text{ ref/verw })=\left(E_{mech}\right)_{Top}/Bo\\ \left(E_{P}+E_{K}\right)_{3m(ref/verw)}=\left(E_{P}+E_{K}\right)_{Top/Bo}\\ \left(mgh+1/2mv^2\right)_{3m(ref/verw)}=\left(mgh+1/2m^2\right)_{Ibp/Bo}\end{array}\right\}\text{ Any one }

(0,06)(9,8)(3)+1/2(0,06)(10)^2=(0,06)(9,8)(h)+0

\mathrm{h}=8,1\mathrm{~m}

Or

Option 9

\left.\begin{array}{l}W_{nc}=\Delta E_{p}+\Delta E_{k}\\ 0=mg\left(h_{f}-h_{i}\right)+1/2m\left(v_{f}^2-v_{i}^2\right)\end{array}\right\}Anyone

0=(0,06)(9,8)\left(h_{f}-3\right)+1/2(0,06)\left(0-10^2\right)

\mathrm{h}=8,1\mathrm{~m}

Or

Option 10

Upwards as positive:

v_{f}^2=v_{i}^2+2a\Delta y

=(10)^2+2(-9,8)(-3)

v_{f}=-12,60

v_{f}=12,6m\cdot s^{-1}(12,62)

v_{f}^2=v_{i}^2+2a\Delta y

(-12,60)^2=(0)^2+2(-9,8)\Delta y

\Delta y=-8,1m

\Delta y=8,1m(8,102-8,103)

Or

Downwards as positive:

v_{f}^2=v_{i}^2+2a\Delta y

=(-10)^2+2(9,8)(3)

\mathrm{v}_{\mathrm{f}}=12,60\mathrm{~m}\cdot\mathrm{s}^{-1}(12,62)

v_{f}^2=v_{i}^2+2a\Delta y

(12,60)^2=(0)^2+2(9,8)\Delta y

\Delta y=8,1m(8,102-8,103)

v_{f}^2=v_{i}^2+2a\Delta y

=(-10)^2+2(-9,8)(-3)

v_{f}=-12,60

v_{f}=12,6\mathrm{~m}\cdot\mathrm{s}^{-1}

\mathrm{v}_{\mathrm{f}}=\mathrm{v}_{\mathrm{i}}+\mathrm{a}\Delta\mathrm{t}

-12,60=0+(-9,8)\Delta t

\Delta t=1,29\mathrm{~s}(1,286)

\Delta y=v_{i}\Delta t+1/2a\Delta t^2

=0+1/2(-9,8)(1,29)^2

\Delta y=-8,1m\quad(-8,098)

\mathrm{h}_1=8,1\mathrm{m}(8,098-8,102)

Or

Option 11

Upwards as positive:

v_{f}^2=v_{i}^2+2a\Delta y

=(10)^2+2(9,8)(3)

\mathrm{V}_{\mathrm{f}}=12,6\mathrm{~m}\cdot\mathrm{s}^{-1}

v_{f}=v_{i}+a\Delta t

12,60=0+(9,8)\Delta t

\Delta\mathrm{t}=1,29\mathrm{~s}(1,286)

\Delta v=v_{i}\Delta t+1/2a\Delta t^2

=0+1/2(-9,8)(1,29)^2

\Delta y=-8,1m(-8,098)

h_1=8,1\mathrm{m}(8,098-8,102)

Or

Downwards as positive:

v_{f}^2=v_{i}^2+2a\Delta y

=(10)^2+2(9,8)(3)

\mathrm{V}_{\mathrm{f}}=12,6\mathrm{~m}\cdot\mathrm{s}^{-1}

v_{f}=v_{i}+a\Delta t

12,60=0+(9,8)\Delta t

\Delta\mathrm{t}=1,29\mathrm{~s}(1,286)

\Delta y=v_{i}\Delta t+1/2a\Delta t^2

=0+1/2(9,8)(1,29)^2

\Delta y=8,1m(8,098)

\mathrm{h}_1=8,1\mathrm{m}(8,098-8,102)

Or

Option 12

Displacement = Area between the graph and the t - axis.

= Area of a triangle

=1/2\mathrm{bh}

=1/2(1,02)(9,996)

=5,09796\mathrm{~m}

\text{ Height }h=3+5,09796

=8,09796\mathrm{m}(8,09796-8,102)

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